Simplify; express your answer in exponential form. Assume $y\neq 0, z\neq 0$. $\dfrac{{(y^{-3})^{-4}}}{{(y^{-1}z^{-5})^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{-3}}$ to the exponent ${-4}$ . Now ${-3 \times -4 = 12}$ , so ${(y^{-3})^{-4} = y^{12}}$ In the denominator, we can use the distributive property of exponents. ${(y^{-1}z^{-5})^{3} = (y^{-1})^{3}(z^{-5})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{-3})^{-4}}}{{(y^{-1}z^{-5})^{3}}} = \dfrac{{y^{12}}}{{y^{-3}z^{-15}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{12}}}{{y^{-3}z^{-15}}} = \dfrac{{y^{12}}}{{y^{-3}}} \cdot \dfrac{{1}}{{z^{-15}}} = y^{{12} - {(-3)}} \cdot z^{- {(-15)}} = y^{15}z^{15}$.